Simplify the following expression and state the condition under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-4}{2t(3t - 1)} \times \dfrac{2t(3t - 1)}{-4} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $r = \dfrac{ -4 \times 2t(3t - 1) } { 2t(3t - 1) \times -4 } $ $ r = \dfrac{-8t(3t - 1)}{-8t(3t - 1)} $ We can cancel the $3t - 1$ so long as $3t - 1 \neq 0$ Therefore $t \neq \dfrac{1}{3}$ $r = \dfrac{-8t \cancel{(3t - 1})}{-8t \cancel{(3t - 1)}} = -\dfrac{8t}{-8t} = 1 $